Question

16.
Factorise 4b2-28bc+49c2-25a² using suitable identities​

Answer 1

Your Answer:

$\tt 25a^2 - 4b^2 + 28bc - 49c^2 \\\\ \tt = (5a)^2 - (4b^2 - 28bc + 49c^2) \\\\ \tt Evaluating\:\: (4b^2 - 28bc + 49c^2) by\:\: \\\\ \tt \star a^2-2ab+b^2=(a-b)^2 \\\\ \tt = (5a)^2 - [{(2b)^2 - 2(2b)(7c) + (7c)^2}] \\\\ \tt Evaluating \:\:by\:\: the\:\: formula \:\:of\:\: \\\\ \tt \star a^2-b^2=(a+b)(a-b) \\\\ \tt = (5a)^2 - (2b - 7c)^2 \\\\ \tt = (5a - 2b + 7c)(5a + 2b - 7c)$

So, factors are (5a-2b+7c) and (5a+2b-7c)

Algebraic Identities used

$\star a^2+b^2+2ab = (a+b)^2 \\\\ \star a^2+b^2-2ab = (a-b)^2 \\\\ \star a^2-b^2=(a+b)(a-b)$

Answer 2

Question:

Factorise 4b²-28bc+49c²-25a² using suitable identities.

Answer:

$The \: factors \: are \: \underline\bold{ (5a + 2b - 7)(5a - 2 - 7c)}$

Step-by-step explanation:

$4 {b}^{2} - 28bc + 49 {c}^{2} - 25 {a}^{2}$

It also can be written as,

$\Rightarrow 25 {a}^{2} - 4 {b}^{2} - 49 {c}^{2} + 28bc$

Now rationalising,

(a - b)² = a² - 2ab + b²

- 4b²- 49c² + 28bc = 4b² - 28bc + 49c²

4b² - 28bc + 49c² = (2b)² - (2)(2b)(7c) + (7c)²

- 4b²- 49c² + 28bc = (2b - 7c)²

$\Rightarrow {(5a)}^{2} - {(2b - 7c)}^{2}$

It is of the form,

a² - b² = (a + b)( a - b)

$\Rightarrow \boxed{(5a + 2b - 7)(5a - 2 - 7c)}$

$The \: factors \: are \: \underline\bold{ (5a + 2b - 7)(5a - 2 - 7c)}$

Identities used:

${(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2}$

$( {a}^{2} - {b}^{2} ) = (a + b)(a - b)$